Question

In: Physics

A car accelerates uniformly from rest and reaches a speed of 23.0 m/s in 8.96 s....

A car accelerates uniformly from rest and reaches a speed of 23.0 m/s in 8.96 s. Assume the diameter of a tire is 57.9 cm.

(a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs.

(b) What is the final angular speed of a tire in revolutions per second?
rev/s

Solutions

Expert Solution

Initial speed of the car = V₁ = 0 m/s

Final speed of the car = V₂ = 23 m/s

Time period = T = 8.96 sec

Diameter of the tire = D = 57.9 cm = 0.579 m

Radius of the tire = R = D/2 = 0.2895 m

Initial angular speed of the tire = ₁ = 0 rad/s

Final angular speed of the tire = ₂

₂ = V₂/R

₂ = 23/0.2895

₂ = 79.447 rad/s

Angular acceleration of the tire =

₂ = ₁ + T

79.447 = 0 + (8.96)

= 8.867 rad/s²

Angle turned by the tires =

= ₁T + T²/2

= (0)(8.96) + (8.867)(8.96)²/2

= 355.928 rad

Number of revolutions made by the tire = n

n = /2

n = 355.928/2

n =56.648 revolutions

₂ = 79.447 rad/s

Converting to rev/s

₂ = 79.447 x (1/2) rev/s

₂ = 12.644 rev/s

a) Number of revolutions the tires make in this motion = 56.648

b) Final angular speed of the tire = 12.644 rev/s

genius_generous answered 1 month ago

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